4 Some properties of bi-Lipschitz mappings

, k = 1, . . . , n−1

and apply each time Lemma 3.1. After at most n−1 steps we get a matrix

C = (c_ij)∈M(n), detC = detA c_jn=c_nj = 0, j = 1, . . . , n−1, c_nn 6= 0

Z

Rⁿ

f(Cx)dx= Z

Rⁿ

f(Ax)dx

This procedure we apply step by step to all further columns and rows. After at most n² steps we find the desired diagonal matrixD with (3.10). By (1.11)

Z

Rⁿ

f(y)dy=|detnD|

Z

Rⁿ

f(Dx)dx=

=|detnA|

Z

Rⁿ

f(Ax)dx.

Combining last identity with (3.9) yields (3.8).

4 Some properties of bi-Lipschitz mappings

Throughout this chapter, let G ⊂ Rⁿ be an open set and, if not otherwise stated, letu:G→Rⁿ be abi-Lipschitz mapping. Let

N ⊂G, |N|= 0

such thatu is differentiable onG\N (compare Theorem 1.5).

Lemma 4.1 Let x₀ ∈ G \ N. Then there is δ = δ(x₀, G, N, u) > 0 such that B_δ(u(x₀))⊂u(G).

Proof.

a) By Theorem 1.5 there is ε⁰ >0 such that Bε⁰(x0)⊂G and there is ϕ:Bε⁰(0)→Rⁿ, lim

Bε0(0)3ξ→0 ξ6=0

ϕ(ξ) kξk = 0 such that

(4.1) u(x) = u(x0) +u⁰(x0)(x−x0) +ϕ(x−x0) ∀x∈Bε(x0).

We choose 0< ε < ε⁰ such that

(4.2) kϕ(ξ)k ≤ L₁

4 kξk ∀ξ∈B_ε(0)

(where 0 < L₁ ≤ L₂ by Theorem 1.6). We set δ := ^L₄¹ε > 0 and for y ∈ B_δ(u(x₀))

(4.3)

u(x, t) :=tu(x) + (1−t) [u(x0) +u⁰(x0)(x−x0)]−y for (x, t)∈Bε(x0)×I.

It is immediately seen that hypotheses (i) and (ii) of Theorem 2.5 hold (with G replaced by B_ε(x₀)). Forkx−x₀k=ε, t∈[0,1] by (4.1)

u(x, t) = u⁰(x0)(x−x0) +tϕ(x−x0) +u(x0)−y By (1.29) and (4.2)

(4.4)







ku(x, t)k ≥ ku⁰(x0)(x−x0)k − kϕ(x−x0)k − ku(x0)−yk

≥L1·ε− ^L₄¹ε− ^L₄¹ε= ^L₂¹ε = 2δ >0

∀kx−x0k=ε, ∀t∈I Let U :={y∈Rⁿ :kyk> δ}. Thenu(x, t)∈U for (x, t)∈∂Bε(x0)×I.

By Theorem 2.5 (applied to B_ε(x₀)) for f ∈C_c⁰(Rⁿ) with f¯

¯U = 0 Z

Bε(x0)

f(u(x)−y) det_nu⁰(x)dx= (4.5)

= Z

Bε(x0)

f(u(x₀) +u⁰(x₀)(x−x₀)−y)·det_nu⁰(x₀)dx

b) Assume now that there is no x∈B_ε(x₀) such that u(x) =y. Then min

n

ku(x)−yk¯

¯x∈B_ε(x₀) o

=:σ > 0.

For x=x0 we see because of y∈Bδ(u(x0))

(4.6) σ ≤ ku(x₀)−yk< δ.

Let

(4.7) ϕ(t) :=

(¡_σ

2 −t¢

for 0≤t ≤ ^σ₂ 0 for t≥ ^σ₂

and f(z) := ϕ(kzk) for z ∈ Rⁿ. f ∈ C_c⁰(Rⁿ) and because of (4.6) it vanishes on U. Further, f(u(x)−y) = 0 for x ∈ B_ε(x₀) and therefore the integral at the left of (4.5) is zero.

On the other hand, det_nu⁰(x₀) 6= 0, and there exists a unique z ∈ Rⁿ such that y−u(x0) = u⁰(x0)·z. By (1.29)

L1ε

4 =δ >ky−u(x0)k=ku⁰(x0)·zk ≥L1kzk, hence kzk< ₄^ε and

x₁ :=x₀+z ∈B_ε/4(x₀),

f(u(x₀) +u⁰(x₀)(x₁ −x₀)−y) =f(0) = σ 2. By continuity there is τ > 0 such that B_τ(x₁)⊂B_ε(x₀) and

f(u(x0) +u⁰(x0)(x−x0)−y)≥ σ

4 ∀x∈Bτ(x1).

Then

|detnu⁰(x0)|

Z

Bε(x0)

f(u(x0) +u⁰(x0)(x−x0)−y)dx≥

≥ σ

4|Bτ(x1)| |detnu⁰(x0)|>0, a contradiction.

Corollary 4.2 Let N^∗ := u(N) ⊂ u(G), |N^∗| = 0 (Lemma 1.7). Then there are open sets V, V^∗ ⊂Rⁿ such that

u(V) =V^∗

G\N ⊂V ⊂G, u(G)\N^∗ ⊂V^∗ ⊂u(G).

Proof. For x∈G\N let Bδx(x) be according Lemma 4.1 and set V^∗ := [

x∈G\N

B_δx(x)

Then V^∗ ⊂ Rⁿ is open, V^∗ ⊂ u(G) and V :=u⁻¹(V^∗)⊂ G is open too. Clearly by construction,G\N ⊂V and because u is injective, u(G)\N^∗ ⊂V^∗.

Lemma 4.3 For each x₀ ∈G\N there is δ =δ(x₀, G, N, u)>0 such that i) B_δ(u(x₀))⊂u(G)

ii) sgn detnu⁰(x) = sgn detnu⁰(x0) for all x∈u⁻¹(Bδ(u(x0))\N) iii) for every f ∈C_c⁰(Rⁿ) with suppf ⊂B_δ(u(x₀))

(4.8)

Z

u(G)

f(y)dy= Z

G

f(u(x))|det_nu⁰(x)|dx

Proof.

a) We proceed similarly as in the proof of Lemma 4.1. We choose ε > 0 and δ = ^L₄^1ε as there. Now we consider

u(x, t) := tu(x) + (1−t) [u(x0) +u⁰(x0)(x−x0)] ∀(x, t)∈Bε(x0)×I.

Again hypotheses i) and ii) of Theorem 2.5 hold with G replaced by B_ε(x₀).

By (4.1)

u(x, t) = u(x₀) +u⁰(x₀)(x−x₀) +tϕ(x−x₀).

For kx−x₀k=ε,t ∈I, by (1.29) and (4.2)

(4.9) ku(x, t)−u(x₀)k ≥L₁kx−x₀k − L₁

4 kx−x₀k= 3

4L₁·ε= 3δ Let now

U⁰ :=©

y ∈Rⁿ¯

¯ky−u(x₀)k> δª By (4.9)

u(x, t)∈U⁰ for (x, t)∈∂B_ε(x₀)×I

and by Theorem 2.5 for all f ∈C_c⁰(Rⁿ) with suppf ∈B_δ(u(x₀))

(4.10)Z for (4.11). Suppose in addition, that h≥ 0, then f ≥0 too. We consider the case σ= 1. Then by (4.11)

Since |detu⁰(x)| ≤C ∀x∈Ω, Z

Ω

|χ_M(x)− |h_k(x)|| |det_nu⁰(x)|dx≤

≤C Z

Ω

|χ_M(x)− |h_k(x)||dx.

By (4.12) Z

Ω\N

χ_M(x) det_nu⁰(x)dx= lim

k→∞

Z

Ω\N

|h_k(x)|det_nu⁰(x)dx≥0

On the other hand χ_M(x) det_nu⁰(x) ≤ 0, for x ∈ Ω\N. Therefore |M| = 0 and detnu⁰(x)>0 a. e. in Ω\N, σ·detnu⁰(x) =|detnu⁰(x)|and (4.7) follows from (4.11). The case σ=−1 is treated analogously.

Lemma 4.4 Let G ⊂ Rⁿ be in addition connected. Then either det_nu⁰(x) > 0 or det_nu⁰(x)<0 ∀x∈G\N.

Proof. Let x_i ∈ G \ N(i = 0,1) x₀ 6= x₁. Then there is a continuous curve γ :I →G such that γ(i) =xi(i= 0,1). Let

¯ γ :=©

γ(t)¯¯t∈[0,1]ª .

Then ¯γ ⊂ G is compact, hence d := dist(¯γ, ∂G) >0 (if ∂G 6= φ; set d := 1 if G = Rⁿ). We choose δ_i > 0 according Lemma 4.3 such that B_δi(x_i) ⊂ u(G) (i = 0,1).

Let

δ:= 1 2min

µ

δ₁, δ₂, d,kx0−x1k 2

¶

>0 Then

G⁰ :=©

x∈G¯

¯dist(x,γ)¯ < δª is a bounded open set subset ofG, G⁰ ⊂⊂G. We set

u(x, t) := u(x)−u(γ(t)) for (x, t)∈G¯⁰×I.

Clearly, ∂G⁰ ⊂ ©

x∈G¯

¯dist(x,γ) =¯ δª

. By compactness of ¯γ for every x ∈ ∂G⁰ there exists t₀ ∈I such that

kx−γ(t0)k=δ= inf©

kx−γ(t)k¯

¯t∈Iª . Then fort ∈I,x∈∂G⁰

(4.13) ku(x, t)k=ku(x)−u(γ(t))k ≥L₁kx−γ(t)k ≥L₁kx−γ(t₀)k=L₁·δ

Let

U :=

½

y∈Rⁿ¯¯dist(y, u(¯γ))> L₁δ 2

¾

By (4.13) u(∂G⁰×I)⊂U. Let nowϕ∈C_c⁰(Rⁿ), suppϕ⊂B_δ(0) and Z

Bδ(0)

ϕ(y)dy = 1.

By Theorem 2.5 (applied toG⁰) Z

G⁰

ϕ(u(x,0)) det_nu⁰(x,0)dx = Z

G⁰

ϕ(u(x,1)) det_nu⁰(x,1)dx that is

Z

G⁰

ϕ(u(x)−u(x₀)) det_nu⁰(x)dx = Z

G⁰

ϕ(u(x)−u(x₁)) det_nu⁰(x)dx

Sinceϕ(u(x)−u(x_i)) vanishes forx /∈u⁻¹(B_δ(u(x_i)), the domain of integration may be replaced by these sets respectively. According Lemma 4.3 ii)

sgn det_nu⁰(x) = sgn det_nu⁰(x_i) =:σ_i ∀x∈u⁻¹(B_δu(x_i))\N, i= 1,2 Therefore

σ₀ Z

u⁻¹(Bδ(u(xo)))

ϕ(u(x)−u(x₀))|det_nu⁰(x)|dx=

=σ₁ Z

u⁻¹(Bδ(u(x1)))

ϕ(u(x)−u(x₁))|det_nu⁰(x)|dx

We apply Lemma 4.3 to each of this integrals and to the mapsvi(x) :=u(x)−u(xi) and we get

σ₀ Z

Bδ(0)

ϕ(y)dy =σ₁ Z

Bδ(0)

ϕ(y)dy

and thereforeσ0 =σ1. Since xi ∈G\N (i= 0,1) had been arbitrary (x0 6=x1), the claim is proved.

The next result is a special case of a famous theorem by L. E. J. Brouwer. This theorem guarantees that a continuous, locally injective mappingf :G→Rⁿ, where G⊂ Rⁿ is open, is an open mapping, i. g. it maps open subsets U ⊂ G onto open subsets f(U)⊂Rⁿ. Especially, f(G) is open. Here, locally injective means that to every x ∈ G there exists a neighborhood U_x ⊂ G such that f¯

¯Ux : U_x → f(U_x) is injective (see e.g. [Dei], Theorem 4.3).

Clearly, locally bi-Lipschitz mappings are continuous and locally injective.

Theorem 4.5 Let G ⊂ Rⁿ be open and let u : G → Rⁿ be a locally bi-Lipschitz mapping. Then u is open.

Proof.

a) We prove first that for each x ∈G there is ε_x >0 such that B_εx(x) ⊂G and there is δ_x > 0 such that B_δx(u(x)) ⊂ u(B_εx(x)). Without loss of generality let x₀ = 0∈G and u(0) = 0. Otherwise consider

˜

u(x) := u(x−x₀)−u(x₀) for x ∈ G˜ := ©

y+x₀¯¯y∈Gª

. Then there is ε > 0 such that B_ε(0) ⊂ G and u¯

¯Bε(0) is bi-Lipschitz (compare Definition 1.3, part 4). Clearly, by continuity, there is a unique bi-Lipschitz extension to B_ε(0) : 0< L₁ ≤L₂

(4.14) L1kx−x⁰k ≤ ku(x)−u(x⁰)k ≤L2kx−x⁰k ∀x, x⁰ ∈Bε(0) Consider y∈B_δ(0), where δ:= ^L₂^1ε >0 and consider

u(x, t) := u(x)−ty for (x, t)∈Bε(0)×I.

Then for (x, t)∈∂B_ε(0)×I by (4.14) and u(0) = 0

(4.15) ku(x, t)k ≥ ku(x)k − kyk ≥L₁kxk − kyk> L₁·ε− L₁ε

2 = L₁ε 2 =δ Suppose that y /∈u

³ B_ε(0)

´

. Then σ:= min

n

ku(x)−yk¯

¯x∈B_ε(0) o

>0.

Clearly,

σ ≤ ku(0)−yk=kyk< δ = L1ε 2 .

Consider again ϕ defined by (4.7) and f(z) :=ϕ(kzk) forz ∈Rⁿ. Let U :=©

y∈Rⁿ¯

¯kyk> δª . By (4.15) u(∂B_ε(0)×I)⊂ U, f ∈ C_c⁰(Rⁿ) and f¯¯

U = 0. By Theorem 2.5 we see

(4.16)

Z

Bε(0)

ϕ(ku(x)−yk) detu⁰(x) = Z

Bε(0)

ϕ(ku(x)k) detu⁰(x)dx

Then ϕ(ku(x)−yk) = 0 ∀x ∈ B_ε(0) and the left integral vanishes. On the other hand u(0) = 0, therefore ϕ(ku(0)k) = ^σ₂. Then there is 0< ε⁰ < ε such that

ϕ(ku(x)k)≥ σ

4 >0 ∀x∈B_ε⁰(0).

Since B_ε(0) is a domain, by Lemma 4.4 det_nu⁰(x) is of constant sign on Bε(0)\N. Without loss of generality let sgn detnu⁰(x) = 1 forx∈Bε(0)\N. By (4.16)

(4.17) 0 =

Z

Bε(0)

ϕ(ku(x)k) detnu⁰(x)dx≥ σ 4

Z

B_ε0(0)

detnu⁰(x)dx.

Because of the first inequality in (1.19), det_nu⁰(x) 6= 0 ∀x ∈ B_ε⁰(0)\N. If we suppose R

B_ε0(0)

det_nu⁰(x)dx = 0, we would conclude (observe det_nu⁰(x) ≥ 0 inR B_ε⁰(0)) det_nu⁰(x) = 0 a. e. in B_ε⁰(0), a contradiction. Therefore

B_ε0(0)

det_nu⁰(x)dx > 0 and we get by (4.17) a contradiction. Since y ∈ B_δ(0) was arbitrary we see B_δ(0) ⊂B_δ(0)⊂u(B_ε(x)).

b) By part a) of proof, for everyx∈Gthere isδx>0 such thatBδx(u(x))⊂u(G).

Then S

x∈G

B_δx(u(x)) is open in Rⁿ and clearly u(G)⊂ [

x∈G

B_δx(u(x))⊂u(G).

Therefore u(G) is open in Rⁿ.

c) If G⁰ ⊂ G is any open set, then we apply parts a) and b) of proof to G⁰ in place of G and we see by b) that u(G⁰) is open in Rⁿ.

Theorem 4.6 Let G ⊂ Rⁿ be open and let u : G → Rⁿ be a locally bi-Lipschitz mapping. Then

i) G^∗ :=u(G) is open

ii) For every x0 ∈G there are open neighborhoods Ux0 ⊂ G and Vy0 ⊂G^∗(y0 :=

u(x₀)) such that u¯

¯Ux0 : U_x0 → V_y0 is bijective and there is a set M ⊂ V_y0,

|M|= 0, such that

v :Vy0 →Ux0, v :=

³ u¯

¯Ux0

´₋₁

is differentiable on V_y0 \M.

Proof. Due to Theorem 4.5, G^∗ is open. There is a neighborhood Ux0 such that u¯

¯Ux0 : U_x0 → Rⁿ is a bi-Lipschitz mapping, hence u : U_x0 → V_y0 := u(U_x0) is bijective andV_y0 is open by Theorem 4.5. Due to Rademacher’s theorem (Theorem 1.4) there isM ⊂V_y0, |M|= 0 such that v :V_y0 \M →U_y0 is differentiable.

Theorem 4.6 is a generalization of the classical theorem on local diffeomorphisms.

Theorem 4.7 Let G ⊂ Rⁿ be open and let u : G → Rⁿ be an injective, locally bi-Lipschitz mapping. Then

i) G^∗ :=u(G) is open,

u:G→G^∗ is bijective and differentiable on G\N, |N|= 0.

ii) There is a setM ⊂G^∗, |M|= 0 such thatv :=u⁻¹ is differentiable onG^∗\M. Proof. By Theorem 4.5, G^∗ is open and because of global injectivity, u :G→G^∗ is bijective. Let (G^∗_k)⊂G^∗ be an exhausting sequence for G^∗:

G^∗_k ⊂⊂G^∗_k+1 ⊂⊂G^∗ ∀k ∈N, [∞

k=1

G^∗_k =G^∗

Due to Theorem 4.6 for each y0 ∈G^∗_k there is an open Vy0 and a set My0, |My0|= 0 such that v¯

¯Vy0\M is differentiable. By compactness of G^∗_k there are y^(k)_i ∈ G^∗_k (i = 1, . . . , m_k) such that G^∗_k ⊂ ^mS^k

i=1

V_y(k)

i , M_i^(k) ⊂ Vy^(k)i , |M_i^(k)| = 0 and v¯

¯V_yi^(k)\M_i^(k) is differentiable. Then v¯

¯G^∗_k\Mk is differentiable, whereMk := ^mS^k

i=1

M_i^(k), |Mk| = 0. Let M := S^∞

k=1

M_k ⊂G^∗,|M|= 0. Then v is differentiable onG^∗\M. As a further corollary we derive

Theorem 4.8 (implicit function theorem) Letm, n∈Nand letU_m ⊂R^m, U_n⊂Rⁿ be open sets, U :=U_m×U_n ⊂R^n+m. Let

f :U →Rⁿ and suppose that

i) there is a∈U_m, b∈U_n such that

f(a, b) = 0 ii) there exists L >0 such that

kf(x, y)−f(x⁰, y⁰)k_n ≤L(kx−x⁰k_m+ky−y⁰k_n) ∀x, x⁰ ∈U_m, ∀y, y⁰ ∈U_n

iii) there exists K >0 such that

kf(x, y)−f(x, y⁰)kn ≥Kky−y⁰kn ∀y, y⁰ ∈Un ∀x∈Um. Then there exists an open neighborhoodV_m ⊂U_m of a and a Lipschitz mapping

g :V_m →Rⁿ such that

1⁰ (x, g(x))∈U ∀x∈V_m

2⁰ there is a subset N_m ⊂V_m, |N_m|= 0 and g is differentiable on V_m\N_m 3⁰ g(a) = b and f(x, g(x)) = 0 ∀x∈V_m

4⁰ ©

(x, y)∈V_m×U_n¯

¯f(x, y) = 0ª

=©

(x, g(x))¯

¯x∈V_mª Proof.

a) Let h:U →R^m+n be defined by

hi(x, y) :=αxi for i= 1, . . . , m

(x, y)∈U_m×U_n =U h_m+k(x, y) :=f_k(x, y) for k= 1, . . . , n

where α=£₁

2K²+L²¤¹

2 >0.

For (x, y), (x⁰, y⁰)∈U

kf(x, y)−f(x⁰, y⁰)k_n≥¯¯kf(x, y)−f(x, y⁰)k_n− kf(x, y⁰)−f(x⁰, y⁰)k_n¯¯. Then by ii) and iii) (for ε >0: 2|a·b| ≤εa²+ε⁻¹b²; chooseε:= ¹₂)

kf(x, y)−f(x⁰, y⁰)k²_n ≥(kf(x, y)−f(x, y⁰)k_n− kf(x, y⁰)−f(x, y⁰)k_n)²

≥ 1

2kf(x, y)−f(x, y⁰)k²_n− kf(x, y⁰)−f(x, y⁰)k²_n

≥ 1

2K²ky−y⁰k²_n−L²kx−x⁰k²_m Hence

kh(x, y)−h(x⁰, y⁰)k²_n+m=α²kx−x⁰k²_m+kf(x, y)−f(x⁰, y⁰)k²_n

≥(α²−L²)kx−x⁰k²_m+ 1

2K²ky−y⁰k²_n

≥ 1

2K²k(x, y)−(x⁰, y⁰)k²_n+m

and therefore

kh(x, y)−h(x⁰, y⁰)k_n+m ≥L₁k(x, y)−(x⁰, y⁰)k_n+m, ∀(x, y),(x⁰, y⁰)∈U (4.18)

L₁ := 1 2K ·√

2>0.

kh(x, y)−h(x⁰, y⁰)k_n+m ≤αkx−x⁰k_m+L(kx−x⁰k_m+ky−y⁰k_m) (4.19)

≤L₂k(x, y)−(x⁰, y⁰)k_n+m ∀(x, y),(x⁰, y⁰)∈U where L₂ :=α+ 2L >0.

Because of (4.18), (4.19)h:U →Rⁿis a bi-Lipschitz mapping and by Theorem 4.7 U^∗ :=h(U) is open. Let

v :=h⁻¹ :U^∗ →U and by (4.18), (4.19) for z ∈Rⁿ, ω ∈R^m, (z, ω)∈U^∗

L⁻¹₂ k(z, ω)−(z⁰, ω⁰)kn+m ≤ kv(z, ω)−v(z⁰, ω⁰)kn+m

(4.20)

≤L⁻¹₁ k(z, ω)−(z⁰, ω⁰)k_n+m ∀(z, ω),(z⁰, ω⁰)∈U^∗. b) Since h(a, b) = (αa,0)∈U^∗ and U^∗ is open, there isδ >0 such that

B_δ(αa,0) =©

(z, ω)∈R^n+m¯

¯k(z, ω)−(αa,0)k_n+m < δª

⊂U^∗. Let

Bm :=

½

z ∈R^m¯

¯kz−αakm < δ

√2

¾

Bn:=

½

ω ∈Rⁿ¯

¯kωkn< δ 2

¾

Obviously (αa,0)∈B_m×B_n ⊂U^∗. Let V_m :=

½

x∈R^m¯

¯kx−ak_m< δ α√

2

¾

Observe that x∈Vm if and only if αx∈Bm. We write v =¡

˜ v,v˜˜¢

, where

˜

v := (v₁, . . . , v_m), v˜˜:= (v_m+1, . . . , v_m+n).

(z, ω)∈U^∗ if and only if there is a unique (x, y)∈U such that (z, ω) = h(x, y) = (αx, f(x, y)).

Therefore z =αx,ω =f(x, y), hence

˜

v(z, ω) =x, v(z, ω) =˜˜ y Since h(a, b) = (αa, f(a, b)) = (αa,0),

˜˜

v(αa,0) =b.

If (x, ω)∈Vm×Bn, then (αx, ω)∈Bm×Bn and

(αx, ω) = h(v(αx, ω)) = (αx, f(x,˜˜v(αx, ω)), hence

ω =f(x,v˜˜(αx, ω)) If ω = 0, 0∈B_n, x∈V_m, then

¡x,˜˜v(αx,0)¢

∈U.

We set

(4.21) g(x) := ˜˜v(αx,0) for x∈V_m g is continuous, (x, g(x))∈U and

0 =f(x, g(x)) ∀x∈V_m.

Further, g(a) = ˜˜v(αa,0) = b. If (x, y) ∈ V_m ×U_n ⊂ U and f(x, y) = 0, then h(x, y) = (αx,0) ∈ U^∗, therefore v(αx,0) = (x,v(αx,˜˜ 0)) = (x, g(x)). Then trivially 4⁰ is true. By (4.20) for x, x⁰ ∈V_m

kg(x)−g(x⁰)k_n =k˜˜v(αx,0)−˜˜v(αx⁰,0)k_n≤

≤ kv(αx,0)−v(αx⁰,0)k_n+m ≤L⁻¹₁ kα(x−x⁰)k_m =

=L⁻¹₁ αkx−x⁰k_m

By Rademacher’s theorem there is Nm ⊂ Vm, |Nm|m = 0, such that g is differentiable on V_m\N_m.

Corollary 4.9 Let the hypotheses of Theorem 4.8 hold true. Then there is N ⊂ R^n+m, |N|_n+m = 0 such that f is differentiable on U \N. For (x, y)∈U\N let

(D_yf)(x, y) =





∂f1

∂y1(x, y) · · · _∂y^∂f_n¹(x, y)

... ...

∂fn

∂y1(x, y) · · · ^∂f_∂yⁿ

n(x, y)





Then

(4.22) Kkηkn ≤ k(Dyf)(x, y)ηkn≤Lkηkn ∀η∈Rⁿ, ∀(x, y)∈U \N and

rank (Dyf)(x, y) = n ∀(x, y)∈U\N.

Proof. The proof is completely analogous to the proof of Theorem 1.5

Clearly f and g are a.e. differentiable and f(x, g(x)) = 0 ∀x ∈ Vm too. At points x ∈ V_m\N_m such that (x, g(x))∈/ N ⊂ R^n+m (N denoting the set where f is not differentiable), then the chain rule may be applied tof(x, g(x)). But it may happen that (x, g(x))∈N ∀x∈Vm. A simple example:

LetU :=©

(x, y)∈R²¯

¯|x|< ¹₂,|y|<1ª

f(x, y) :=







(1 +x)y for y≥0

(x, y)∈U (1 +x)2y for y <0

Thenf(0,0) = 0,

kf(x, y)−f(x⁰, y⁰)k₁ ≤3(kx−x⁰k₁+ky−y⁰k₁) kf(x, y)−f(x, y⁰)k₁ ≥ 1

2ky−y⁰k₁ ∀|x|< 1

2 ∀|y|<1.

All hypotheses of Theorem 4.8 hold true. Let V_m :=©

x∈R¯

¯|x|< ¹₂ª

, g : V_m →R, g(x) := 0. Then f(x, g(x)) = 0 ∀x∈V_m. The set where f is not differentiable is

N :=©

(x, y)∈U¯

¯y= 0ª

⊂R², |N|₂ = 0 and ©

(x, g(x))¯

¯x∈V_mª

=N. The chain rule is not applicable.

5 The change of variables formula for (locally)

Im Dokument A homotopy argument and its applications to the transformation rule for bi-Lipschitz mappings, the Brouwer fixed point theorem and the Brouwer degree (Seite 32-45)